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Phantasy [73]
2 years ago
10

What was the original price of a bookshelf whose sales price is $16

Mathematics
1 answer:
sattari [20]2 years ago
5 0
Mi madre se aprende de memoria la dirección de Jimin
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Question 4 (1 point)
igomit [66]

Answer:

x = 15

Step-by-step explanation:

We want to solve for x in  3*(2x + 5) = 3x + 4x

First simplify.

3*(2x + 5) = 7x

Next, distribute the 3.

3*2x + 3*5 = 7x

6x + 15 = 7x

15 = 7x - 6x

15 = x

4 0
3 years ago
Jonas bought 3 books. Each book was the same price. After using a $10-off coupon, the total charge was $20. Which equation can b
Brums [2.3K]

Answer:

3b-10=20

Step-by-step explanation:

3b-10=20

3b-10+10=20+10

3b=30

3b/3=30/3

b=10

8 0
1 year ago
Evaluate 7C3. 35 210 21 105
nydimaria [60]

\bf _nC_r=\cfrac{n!}{r!(n-r)!}\qquad \qquad \qquad \qquad \qquad _7C_3=\cfrac{7!}{3!(7-3)!} \\\\\\ \textit{use the }[!]\textit{ factorial button or the }[_nC_r]\textit{ button in your calculator}

8 0
3 years ago
Solve 11 2/5 = q - 4 2/7 + 2 1/7
RideAnS [48]
I turn the mixed numbers into decimals before doing anything else.
11 2/5----> 57/5----> 11.4
4 2/7----> 30/7----> 4.29
2 1/7----> 15/7----> 2.14

11.4=q-4.29+2.14
11.4=q-2.15
13.55= q

13.55 converted to a fraction is 13 11⁄20
q=13 11⁄20
6 0
3 years ago
Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find
Papessa [141]

Answer:

1) \text{P(at least one boy and one girl)}=\frac{3}{4}

2) \text{P(at least one boy and one girl)}=\frac{3}{8}

3) \text{P(at least two girls)}=\frac{1}{2}

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To  Find : The probability of each event.  

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)        

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}

1) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, BGG, GBB, GBG, GGB=6

\text{P(at least one boy and one girl)}=\frac{6}{8}

\text{P(at least one boy and one girl)}=\frac{3}{4}

2) P(at least one boy and one girl)

Favorable outcome =  BBG, BGB, GBB=3

\text{P(at least one boy and one girl)}=\frac{3}{8}

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

\text{P(at least two girls)}=\frac{4}{8}

\text{P(at least two girls)}=\frac{1}{2}

4 0
3 years ago
Read 2 more answers
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